Question

Given \(\frac{d^2y}{dx^2}=cot\,x\frac{dy}{dx}+4y\,cosec^2x=0\). Changing the independent variable x to z by the substitution z=log tan \(\frac{x}{2}\), the equation is changed to 

• A. \(\frac{d^2y}{dz^2}+\frac{3}{y}=0\)
• B. \(2\frac{d^2y}{dz^2}+e^y=0\)
• C. \(\frac{d^2y}{dz^2}-4y=0\)
• D. \(\frac{d^2y}{dz^2}+4y=0\)

Answer 1

The Correct Option is C

To simplify the given equation using the substitution \( z = \log(\tan(x/2)) \), we follow these steps:

Step 1: Express trigonometric functions in terms of \( \tan(x/2) \): 
Let \( t = \tan(x/2) \), so that \( z = \log(t) \Rightarrow t = e^z \)

Using known identities:
\[ \cot x = \frac{1 - t^2}{2t}, \quad \sec^2(x/2) = 1 + t^2 = 1 + e^{2z}, \quad \csc^2 x = \frac{1 + t^2}{4t^2} \]
Step 2: Chain Rule — Convert derivatives with respect to \( x \) into derivatives with respect to \( z \):
\[ \frac{dy}{dx} = \frac{dy}{dz} \cdot \frac{dz}{dx}, \quad \frac{d^2y}{dx^2} = \frac{d^2y}{dz^2} \left( \frac{dz}{dx} \right)^2 + \frac{dy}{dz} \cdot \frac{d^2z}{dx^2} \]
Using \( z = \log(\tan(x/2)) \), we differentiate:
\[ \frac{dz}{dx} = \frac{d}{dx} \log(\tan(x/2)) = \frac{1}{\tan(x/2)} \cdot \frac{1}{2} \sec^2(x/2) = \frac{1}{2} \cdot \frac{\sec^2(x/2)}{\tan(x/2)} \]
Now express \( \frac{dz}{dx} \) and \( \frac{d^2z}{dx^2} \) in terms of \( z \), then substitute back into the differential equation. After full simplification:

The resulting equation is:
\[ \frac{d^2y}{dz^2} - 4y = 0 \]
Correct Answer: (C)